Check Full Chapter Explained - Continuity and Differentiability - Application of Derivatives (AOD) Class 12       1. Chapter 6 Class 12 Application of Derivatives
2. Serial order wise
3. Miscellaneous

Transcript

Misc 15 Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius r is 4𝑟/3 . Given, Radius of sphere = r Let R be the radius of the cone and H be its height. Let ∠ BOC = θ Now, AC = AO + OC H = r + r cos θ And, R = r sin θ H = r (1 + cos θ) We need to maximize volume of cone. Volume of the cone is V = 1/3 𝜋𝑅^2 𝐻 V = 1/3 𝜋𝑟^2 sin^2⁡"θ" r (1 + cos θ) V = 1/3 𝜋𝑟^3 sin^2⁡"θ" (1 + cos θ) Finding 𝑑𝑣/𝑑𝜃 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 [2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )+sin^2⁡〖"θ" (−sin⁡〖"θ" )〗 〗 〗 〗 〗 ] 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )−sin^3⁡"θ" 〗 〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 (2 sin⁡〖"θ" cos⁡〖"θ" (1+cos⁡〖"θ" )−sin^3⁡"θ" 〗 〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos⁡〖"θ" +〖2 cos^2〗⁡〖"θ" −sin^2⁡"θ" 〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (2 cos⁡〖"θ" +〖2 cos^2〗⁡〖"θ" −(1−cos^2⁡"θ" 〗 〗)) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 cos^2⁡〖"θ" +〖2 𝑐𝑜𝑠〗⁡〖"θ" −1〗 〗 ) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 cos^2⁡〖"θ" +〖3 𝑐𝑜𝑠〗⁡𝜃 〗 −cos⁡𝜃 −1) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 〖𝑐𝑜𝑠 〗⁡"θ" (cos⁡𝜃+1)−1(cos⁡𝜃+1)) 𝑑𝑣/𝑑𝜃 = 1/3 𝜋𝑟^3 sin θ (3 𝑐𝑜𝑠⁡〖"θ" −1〗 )(𝑐𝑜𝑠⁡〖"θ" +1〗 ) Putting 𝑑𝑣/𝑑𝜃 = 0 1/3 𝜋𝑟^3 sin θ (𝑐𝑜𝑠⁡〖"θ" +1〗 ) (3 𝑐𝑜𝑠⁡〖"θ" −1〗 ) = 0 sin θ (cos θ + 1) (3 cos "θ" − 1) = 0 𝐬𝐢𝐧 θ = 0 θ = 0° θ cannot be 0° for cone 𝐜𝐨𝐬 θ + 1 = 0 cos θ = −1 For cone, 0° < θ < 90° & cos θ is negative in II & III quadrant. So cos θ = −1 is not possible 𝟑 𝐜𝐨𝐬 θ − 1 = 0 cos θ = 1/3 θ = cos−1 1/3 cos θ = 1/3 is possible So, 𝑐𝑜𝑠⁡"θ" = 1/3 Thus, H = r (1 + cos θ) H = r ("1 + " 1/3) H = 4𝑟/3 This value is either maxima or minima. So, finding (𝑑^2 𝑉)/(𝑑𝜃^2 ) 𝑑𝑉/𝑑𝜃 = (1/3 𝜋𝑟^3 sin⁡𝜃 (3 cos2 𝜃 + 2 cos⁡𝜃− 1)) = 1/3 𝜋𝑟^3 [cos⁡𝜃 (3 cos^2⁡𝜃+2 cos⁡𝜃−1)+sin⁡𝜃 (6 cos⁡𝜃 (−𝑠𝑖𝑛)−2 sin⁡𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^3⁡𝜃+2 cos^2⁡𝜃−cos⁡𝜃+sin⁡𝜃 (−6 cos⁡𝜃 sin⁡𝜃−2 sin⁡𝜃 )] = 1/3 𝜋𝑟^3 [3 cos^2⁡𝜃+2 cos^2⁡𝜃−cos⁡𝜃−6 sin^2⁡𝜃 cos⁡𝜃−2 sin^2⁡𝜃 ] Now, cos⁡𝜃=1/3 And sin2 "θ" = 1 − cos2 "θ" = 1 − (1/3)^2= 1 − 1/9 = 8/9 Putting values in (𝑑^2 𝑉)/(𝑑𝜃^2 ) (𝑑^2 𝑉)/(𝑑𝜃^2 ) = 1/3 𝜋𝑟^3 [3(1/3)^3+2(1/3)^2−1/3−6(1/3)(8/9)−2(8/9)] = 1/3 𝜋𝑟^3 [3(1/27)+2(1/9)−1/3−6(8/27)−2(8/9)] = 1/3 𝜋𝑟^3 [1/9+ 2/9−1/3−16/9−16/9] = 1/3 𝜋𝑟^3 [(−32)/9] = (−32𝜋𝑟^3)/27 Thus, (𝑑^2 𝑣)/(𝑑𝜃^2 ) < 0 H = 𝟒𝒓/𝟑 Hence proved

Miscellaneous 